I am working on a nodejs/express project which have a deprecated way of accessing request parameters :
req.param('parameter') instead of the valid req.params.parameter.
There is a lot of occurence in the project and I need to fix it. My question is how can I do to find the string's start "req.param('" until the string's end "')", extract parameter from the result, and then replace to have req.params.parameter ?
find:
req.param('str')
replace:
req.params.str
NOTE :
Following the validated answer, to reverse this replace pattern, use :
(req\.params\.)([^; \n]+)
In the find and replace box (ctrl+h) you can insert as the search regex:
(req\.param)\('([^']+)'\)
and replace it with
$1.$2
Explanation:
The first group (req\.param) is accessed via $1 and you could also change it to match more prefixes. Afterwards open a paren and a string. In the second group ([^']+) everything except a string closing char is matched. And afterwards the string and the paren is closed.
Related
I am trying to use Notepad ++ to delete emails that end in #domain2.serverdata.net
here is a string example:
smtp:name#domain1.com;SMTP:name#domain2.com;smtp:name#domain2.serverdata.net;smtp:name#domain3.com;smtp:name_e4d1fe3d-e985-40d0-bc65-32c57c9b14d1#domain2.serverdata.net
I was hoping to use:
;smtp:.*#domain2.serverdata.net
but it captures SMTP:name#domain2.com as well
Ctrl+H
Find what: (?:\A|;)smtp:[^#]*#domain2\.serverdata\.net
Replace with: LEAVE EMPTY
check Wrap around
check Regular expression
Replace all
Explanation:
(?:\A|;) # non capture group, beginning of file or semicolon,
this allows to delete the first email of the file
that haven't a semicolon before it
smtp: # literally
[^#]+ # 1 or more any character that is not #
#domain2\.serverdata\.net # literally
Try Regex: ;?smtp:[\w.-]+?#domain2\.serverdata\.net
Demo
Regexes will usually capture as much as possible. For instance: START.*STOP applied to the following text:
STARTsghlegdSTOPfsgikbSTARTsvdinusSTOPwegtgw
will capture this part:
STARTsghlegdSTOPfsgikbSTARTsvdinusSTOPwegtgw
^------------------------------------^
In your case, the .* captures everything up to the last instance of #domain2.serverdata.net. You don't want to use . (any character), you want to use "any character except '#'" which is written like this: [^#].
So your full regex would be smtp:[^#]*#domain2\.serverdata\.net. I also dropped the initial ; since it would prevent you from capturing the first mail address.
Try this one:
smtp:[^#]+#domain2\.serverdata\.net(;)?
I have multiple occurance of src={icons.ICON_NAME_HERE} in my code, that I would like to change to name="ICON_NAME_HERE".
Is it possible to do it with regular expressions, so I can keep whatever is in code as ICON_NAME_HERE?
To clarify:
I have for example src={icons.upload} and src={icons.download}, I want to do replace all with one regexp, so those gets converted to name="upload" and name="download"
Try searching on the following pattern:
src=\{icons\.([^}]+)\}
And then replace with your replacement:
name="$1"
In case you are wondering, the quantity in parentheses in the search pattern is captured during the regex search. Then, we can access that captured group using $1 in the replacement. In this case, the captured group should just be the name of the icon.
I don't know anything about Notepad++ Regex.
This is the data I have in my CSV:
6454345|User1-2ds3|62562012032|324|148|9c1fe63ccd3ab234892beaf71f022be2e06b6cd1
3305611|User2-42g563dgsdbf|22023001345|0|0|c36dedfa12634e33ca8bc0ef4703c92b73d9c433
8749412|User3-9|xgs|f|98906504456|1534|51564|411b0fdf54fe29745897288c6ad699f7be30f389
How can I use a Regex to remove the 5th and 6th column? The numbers in the 5th and 6th column are variable in length.
Another problem is the User row can also contain a |, to make it even worse.
I can use a macro to fix this, but the file is a few millions lines long.
This is the final result I want to achieve:
6454345|User1-2ds3|62562012032|9c1fe63ccd3ab234892beaf71f022be2e06b6cd1
3305611|User2-42g563dgsdbf|22023001345|c36dedfa12634e33ca8bc0ef4703c92b73d9c433
8749412|User3-9|xgs|f|98906504456|411b0fdf54fe29745897288c6ad699f7be30f389
I am open for suggestions on how to do this with another program, command line utility, either Linux or Windows.
Match \|[^|]+\|[^|]+(\|[^|]+$)
Repalce $1
Basically, Anchor to the end of the line, and remove columns [-1] and [-2] (I assume columns can't be empty. Replace + with * if they can)
If you need finer detail then that, I'd recommend writing a Java or Python script to manual parse and rewrite the file for you.
I've captured three groups and given them names. If you use a replace utility like sed or vimregex, you can replace remove with nothing. Or you can use a programming language to concatenate keep_before and keep_after for the desired result.
^(?<keep_before>(?:[^|]+\|){3})(?<remove>(?:[^|]+\|){2})(?<keep_after>.*)$
You may have to remove the group namings and use \1 etc. instead, depending on what environment you use.
Demo
From Notepad++ hit ctrl + h then enter the following in the dialog:
Find what: \|\d+\|\d+(\|[0-9a-z]+)$
Replace with: $1
Search mode: Regular Expression
Click replace and done.
Regex Explain:
\|\d+ : match 1st string that starts with | followed by number
\|\d+ : match 2nd string that starts with | followed by number
(\|[0-9a-z]+): match and capture the string after the 2nd number.
$ : This is will force regex search to match the end of the string.
Replacement:
$1 : replace the found string with whatever we have between the captured group which is whatever we have between the parentheses (\|[0-9a-z]+)
I've got a file with several (1000+) records like :
lbc3.*'
ssa2.*'
lie1.*'
sld0.*'
ssdasd.*'
I can find them all by :
/s[w|l].*[0-9].*$
What i want to do is to replace the final part of each pattern found with \.*'
I can't do :%s//s[w|l].*[0-9].*$/\\\\\.\*' because it'll replace all the string, and what i need is only replace the end of it from
.'
to
\.'
So the file output is llike :
lbc3\\.*'
ssa2\\.*'
lie1\\.*'
sld0\\.*'
ssdasd\\.*'
Thanks.
In general, the solution is to use a capture. Put \(...\) around the part of the regex that matches what you want to keep, and use \1 to include whatever matched that part of the regex in the replacement string:
s/\(s[w|l].*[0-9].*\)\.\*'$/\1\\.*'/
Since you're really just inserting a backslash between two strings that you aren't changing, you could use a second set of parens and \2 for the second one:
s/\(s[w|l].*[0-9].*\)\(\.\*'\)$/\1\\\2/
Alternatively, you could use \zs and \ze to delimit just the part of the string you want to replace:
s/s[w|l].*p0-9].*\zs\ze\*\'$/\\/
Im trying to use the following regex to search and replace in multiple files in notepad++
([^\n]*)(state="1")([^\n]*)*.
This searches and finds state="1" in the first line of each file and works fine.
However, when I try to replace state="1" using:
Replace with: $1 state="5"
it cuts off the rest of the line.
I thought that it might be possible to get the rest of the line using:
Replace with: $1 state="5" $2
However, $2 doesnt seem to exist as a variable.
Is there some way to attach the rest of the line into variable $2?
Cheers
Heres an image to show how
(?=\A[^\n]*)state="1"
is not working
Ive updated my version of notepad++ and everything
Each capture group, (…), is assigned a number, so $2 represents the second capture group, (state="1"). The remainder of the line is captured in $3.
Either remove the capture group around state="1":
([^\n]*)state="1"([^\n]*)*.
Or use $3:
Replace with: $1 state="5" $3
Also, given the simplicity of the task, I don't see why you couldn't just search for state="1" and replace with state="5". There doesn't seem to be any need for regular expressions here.
Update There's nothing in the pattern listed so far which limits the result to only matching strings on the first line. If you need that I'd recommend using a pattern like this:
(?=\A[^\n]*)state="1"
With these settings:
Update There seems to be some strange behavior with the \A (beginning of text) anchor inside the lookbehind. Removing from the lookbehind seems to work. Try this pattern:
\A([^\n]*)state="1"
And replace with:
$1state="5"
All the other settings should be fine.